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xxMikexx [17]
3 years ago
10

Identify the spectator ions in this reaction. Check all that apply. 2H+ + CrO4– + Ba2+ + 2OH– Ba2+ + CrO4– + 2H2O H+ CrO4– Ba2+

OH– 2H2O
Chemistry
2 answers:
Luba_88 [7]3 years ago
6 0

Answer: B and C on the tenth question of the assignment

bagirrra123 [75]3 years ago
5 0
Answer: CrO₄⁻ and  Ba²⁺


Explanation:


1) Chemical equation given:


2H⁺ + CrO₄⁻ + Ba²⁺ + 2OH⁻ → Ba²⁺ + CrO₄⁻ + 2H₂O


2) Analysis


That is an oxidation-reduction equation (some species are been oxidized and others are being reduced).


The given equation is known as total ionic equation, because it shows all the species as ions that are part of the reaction.


2) Specator ions


Spectator ions are the ions that do not change their oxidation state and are easily identified as they are the same in the reactant and product sides.


Here the ions that are the same in the reactant and product sides are:

CrO₄⁻ and  Ba²⁺


3) Addtitional explanation.


Once you identify the spectator ions you can delete them from the equation to obtain the net ionic equation , which in this case turns to be:


2H⁺ + 2OH⁻ →  2H₂O


But this is  not part of the question; it is some context to help you understand the use of the spectator ions concept.
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Let us consider, volume of air per breathe= x ml.

Pure O_{2} from inhaled air= \frac{20}{100}x ml and Pure O_{2} from exhaled air= \frac{16}{100}x ml.

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A plastic bottle is closed outdoors on a cold day when the temperature is −15.0°C and is later brought inside where the temperat
Lana71 [14]

Answer:

Pressure = 1.14 atm

Explanation:

Hello,

This question requires us to calculate the final pressure of the bottle after thermal equilibrium.

This is a direct application of pressure law which states that in a fixed mass of gas, the pressure of a given gas is directly proportional to its temperature, provided that volume remains constant.

Mathematically, what this implies is

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P1 / T1 = P2 / T2 = P3 / T3 =........= Pn / Tn

P1 / T1 = P2 / T2

P1 = 1.0atm

T1 = -15°C = (-15 + 273.15)K = 258.15K

P2 = ?

T2 = 21.5°C = (21.5 + 273.15)K = 294.65K

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3 0
3 years ago
A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out
scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<u>Explanation:</u>

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol

  • The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = \frac{2}{1}\times 0.0021=0.0042mol of potassium bromide

  • Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g

  • To calculate the percentage composition of KBr in the mixture, we use the equation:

\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%

Hence, the percent by mass of KBr in the mixture is 9.996 %

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